An ebomiote rod (K = 3), 6 mm thick is introduced between the plates of a parallel plate capacitor of plate area `4xx10^(-2) m^(2)` and plate separation 0.01m. Find the capacitance.

Find out the capacitance of parallel plate capacitor of plate area A and plate separation d.

Calculate capacitance of a parallel plate capacitor with area of each plate 1 cm^(2) and separation 1 mm.

A dielectric of thickness 5cm and dielectric constant 10 is introduced between the plates of a parallel plate capacitor having plate area 500 sq. cm and separation between the plates 10cm. The capacitance of the capacitor with dielectric slab is (epsi_(0) = 8.8 xx 10^(-12)C^(2)//N-m^(2))

If a dielectric slab of thickness 5 mm and dielectric constant K=6 is introduced between the plates of a parallel plate air capacitor, with plate separation of 8 mm, then its capacitance is

A dielectric slab of thickness 1.0 cm and dielectric constant 5 is placed between the plates of a parallel plate capacitor of plate area 0.01 m^(2) and separation 2.0 cm. Calculate the change in capacity on introduction of dielectric. What would be on the change, if the dielectric slab were conducting?

A metel plate of thickness 2 cm is introduced between the plates of a parallel plate air capacitor having a plate separation of 6 cm. what is the ratio of the capacities of the capacitor before and after introducing the metal plate?

Area of a parallel plate capacitor of capacitance 2F and separation between the plates 0.5cm will be

The' distance between the plates of a parallel plate capacitor is d . A metal plate of thickness d//2 is placed between the plates. What will e its effect on the capacitance.

The plates of a parallel plate capacitance 1.0 F are separated by a distance d =1 cm. Find the plate area .